Suppose I have $u_{m}$ as a infinite vector. Then the central difference operator $\delta_{0}u_{m}= \frac{u_{m+1}-u_{m-1}}{2h}$ . What is ($\delta_{0})^{2} u_{m}$? Find the order of approximation to ($\delta_{0})^{2} u_{m} - u_{xx}$.
So I found $(\delta_{0})^{2} u_{m}$ = $\delta_{0}(\delta_{0}u_{m})$ = $\delta_{0}w_{m}$ = $\frac{w_{m+1}-w_{m-1}}{2h}$ = $\frac{\frac{u_{m+2}-u_{m}}{2h} - \frac{u_{m}-u_{m-2}}{2h}}{2h}$ = $\frac{u_{m+2}-2u_{m}+u_{m-2}}{4h^{2}}$
Then to compute ($\delta_{0})^{2} u_{m} - u_{xx}$ = $\frac{u_{m+2}-2u_{m}+u_{m-2}}{4h^{2}} - u_{xx}$ I need to expand each $u_{m+2}$ and $u_{m-2}$ using Taylor's Theorem, but when doing so I obtain
($\delta_{0})^{2} u_{m} - u_{xx}$ = $\frac{u_{m+2}-2u_{m}+u_{m-2}}{4h^{2}} - u_{xx}$ = $u_{xx} + O(h^2)$ but I think the $u_{xx}$ should have canceled. What did I do wrong here?
$$u_{m+2} = u_m + 2h u_x(x_m) + \frac 12 (2h)^2 u_{xx}(x_m)+\frac16 (2h)^3 u_{xxx}(x_m)+O(h^4),\\ u_{m-2} = u_m - 2h u_x(x_m) + \frac 12 (2h)^2 u_{xx}(x_m)-\frac16 (2h)^3 u_{xxx}(x_m)+O(h^4),$$ hence
$$u_{m+2}-2u_{m}+u_{m-2} = {u_m + 2h u_x(x_m) + \frac 12 (2h)^2 u_{xx}(x_m)+\frac16 (2h)^3 u_{xxx}(x_m)+O(h^4)+u_m - 2h u_x(x_m) + \frac 12 (2h)^2 u_{xx}(x_m)-\frac16 (2h)^3 u_{xxx}(x_m)+O(h^4)-2u_m}$$
$$ = { \frac 12 (2h)^2 u_{xx}(x_m) O(h^4)+ \frac 12 (2h)^2 u_{xx}(x_m) +O(h^4)} = 4h^2 u_{xx}(x_m)+O(h^4).$$